Ответ: x=0,5 y=1.
Объяснениe:
Пусть:
0;2^{y}=v>0.\\\left \{ {{u-v=1} \atop {\frac{1}{u}-\frac{1}{v} } =-\frac{1}{6} }} \right. ;\left \{ {{u-v=1} \atop {\frac{v-u}{u*v} =-\frac{1}{6} }} \right. ;\left \{ {{u-v=1} \atop {\\\frac{u-v}{u*v} =\frac{1}{6} }} ;\right. \;\left \{ {{u=v+1} \atop {\frac{1}{u*v} =\frac{1}{6} }} \right. ;\left \{ {{u=v+1} \atop {u*v=6}} ;\right.\\\left \{ {{u=v+1} \atop {v*(v+1)=6}} \right. ;\left \{ {{u=v+1} \atop {v^{2}+v-6 =0}} \right. ;\left \{ {{u=v+1} \atop {v^{2}+v+2v-2v-6 =0}} \right. ;" alt="9^{x}=u>0;2^{y}=v>0.\\\left \{ {{u-v=1} \atop {\frac{1}{u}-\frac{1}{v} } =-\frac{1}{6} }} \right. ;\left \{ {{u-v=1} \atop {\frac{v-u}{u*v} =-\frac{1}{6} }} \right. ;\left \{ {{u-v=1} \atop {\\\frac{u-v}{u*v} =\frac{1}{6} }} ;\right. \;\left \{ {{u=v+1} \atop {\frac{1}{u*v} =\frac{1}{6} }} \right. ;\left \{ {{u=v+1} \atop {u*v=6}} ;\right.\\\left \{ {{u=v+1} \atop {v*(v+1)=6}} \right. ;\left \{ {{u=v+1} \atop {v^{2}+v-6 =0}} \right. ;\left \{ {{u=v+1} \atop {v^{2}+v+2v-2v-6 =0}} \right. ;" align="absmiddle" class="latex-formula">
u₁=-2 ∉; v₁=-3 ∉ ⇒